Question

A projectile is fired at 45.0° above the horizontal. Its initial speed is equal to 42.5 m/s. Assume that the free-fall acceleration is constant throughout and that the effects of the air can be ignored. What is the maximum height reached by the projectile?

At what time after being fired does the projectile reach this maximum height?

Answer 1

Vi = 42.5 sin 45 = 30 m/s

v = Vi - 9.81 t
0 = 30 - 9.81 t
t = 3.06 seconds to top

h = 0 + Vi t - (1/2)9.81 t^2
h = 30(3.06) - 4.9 (3.06)^2
= 91.8 - 45.9
= 91.7 m

Answer 2

a ball thrown vertically upward from the top of a tower 60m high a velocity of 30ms what is the maximum height above the ground level.how long does it take to reach the ground